Number System Practice Gym by Anubhav Sehgal  Part 1

Q1) Two numbers when divided by m(m < = 7) gives remainder 3 and 4. The remainder when the sum of these two numbers is divided by the same divisor is R. Find the value of R + m.
Let the two numbers be a and b
a = mq + r1
b = nq + r2
Then, (a + b) = (m + n)q + (r1 + r2)
Remainder when sum of two numbers (giving individual remainders r1 and r2 with a divisor) is divided by the same divisor is :
R = (r1 + r2)  divisor [if this is positive]
else
R = (r1 + r2)
Here R = (3 + 4)  m
R + m = 7Q2) Calculate GCD(1,155) + GCD(2,155) + .. + GCD(155,155). where GCD(m,n) is the greatest common divisor of m and n.
GCD(x,155) = 1 unless x is a multiple of 5 and/or 31
4 onlymultiples of 31
30 onlymultiples of 5
1 multiple of both 31 and 5
35 such numbers
Hence,
Sum = 4 * 31 + 30 * 5 + 1 * 155 + (155  35) * 1 = 549This is an example of how test setters combine two simple concepts to trap you.
Q3) Find the largest number n such that (n  11) divides n^3 + 83.
A mix of elementary algebra and number system
By remainder theorem,
Remainder when n^3 + 83 is divided by (n  11) = 11^3 + 83 = 1414
Now,
(n  11) * k = 1414
For largest n,
k = 1, (n  11) = 1414
n = 1425Q4) Consider natural numbers a, b, c, d and e.Three statements are provided:
I. (de)^2 = abbb
II.(ee)^2 = ccbb
III. A perfect square never ends with c,d and e.Which of the following statements are sufficient to determine the arithmetic mean of a,b and c?
a) Any of the three statements
b) Any two of the three statements
c) Statement I and II are sufficient
d) Statement II and III are sufficient
e) All three statements are necessarySquare with largest occurrence of a nonzero digit at the end is 38^2 = 1444
The only square of the form XX^2 = YYZZ is 88^2 = 7744
A square never ends with 2,3,7,8 or odd number of zeroes
Combine them, analyze the options and you ll find I and II are sufficient to answer what we need which is primarily values of a,b,c.Q5) Find 24^1202 mod 1446.
Q6) Find the smallest whole number that when divided by 5,7,9 and 11 leaves a remainder of 1,2,3 and 4 respectively.
If the difference between divisors and respective remainders was equal you simply apply : LCM(divisors)  common difference. But here you need to tune the formula and that can be done when you know how the standard formula comes.
If N was your original number where
N = 5a + 1 = 7b + 2 = 9c + 3 = 11d + 4
then :Q7) Find the highest N such that 11^N divides (97! + 98! + 99!)
OA : 10, Guess doesn't need discussion
Q8. When a number N(N > 500) is successively divided by 3,5,7, it leaves a remainder of 2,3,6 respectively. Find the remainder when 2nd such smallest N is divided by 35.
N = 3(5(7m + 6) + 3) + 2 = 105m + 101
N = 101,206,311,416,521,626
626 mod 35 = 31Q9. Find the highest power of 1001 in 1001 * 999 * 997 * 995 * ... * 3 * 1
Approach 1 :
Multiply and divide by 2 * 4 * 6 * 8 *... * 998 * 1000
Number = 1001!/(2 * 4 * 6 * 8 * .. * 998 * 1000) = 1001!/2^500 * 500!
Exponent of 13 in this = (77 + 5)  (38 + 2) = 42Approach 2 :
For power of 13, 13 * 1 to 13 * (7 * 11) all odd powers
So, (7 * 11 + 1)/2 = 39
For 13^2,
13 * (13 * 1), 13 * (13 * 3), ... upto 13 * (13 * 5)
So additional three powers.
Total = 39 + 3 = 42Q10) If p, q, and r are positive integers such that (p + q  r)(q + r  p)(p + r  q) = 15, then what is the product of p, q and r?
a) 24
b) 60
c) 64
d) Cannot Be DeterminedOA : CBD
15 can be broken into three factors as follows  (5, 3, 1) or (15, 1, 1).
Solving for p, q, r we get (4, 3, 2) and (8, 8, 1) as the triplets corresponding to the above set of factors.
Thus product of p, q and r can be either 24 or 64.Q11) Sum of two numbers and their LCM is p where p is a prime and 10 < p < 30. For how many p do we have 3 pairs of such number?
Let the numbers be ha and hb.
LCM = hab since LCM * HCF = ha * hb
So,
h(a + b + ab) = p
Case 1 : h != 1
Then, h = p => prime => Not possible
Since LCM >= Numbers >= HCF always
Case 2 : h = 1
a + b + ab = p
(a + 1)(b + 1) = (p + 1)
Number of pairs = [Factors of (p + 1)]/2  1 for the case when a becomes zero.
For the given range,
p = 11, 13, 17, 19, 23, 29
p = 11 => 12 = 2^2 * 3 => 3 * 2/2  1 = 2 pairs
p = 13 => 14 = 2 * 7 => 2 * 2/2  1 = 1 pair
p = 17 => 18 = 2 * 3^2 => 2 pairs
p = 19 => 20 = 2^2 * 5 => 2 pairs
p = 23 => 24 = 2^3 * 3 => 4 * 2/2  1 = 3 pairs
p = 29 => 30 = 2 * 3 * 5 => 2 * 2 * 2/2  1 = 3 pairs
So 2 such p = 23,29Q12) A natural number N is having a total of 21 composite factors. Let X be the maximum number of prime factors of N then find the remainder when X is divided by 3.
X = 2. Hence remainder X mod 3 = 2